Đặt $E=\dfrac{y^4}{(x^2+y^2)(x+y)}+\dfrac{z^4}{(y^2+z^2)(y+z)}+\dfrac{x^4}{(z^2+x^2)(z+x)}$
Ta có:
$F-E=\dfrac{x^4-y^4}{(x^2+y^2)(x+y)}+\dfrac{y^4-z^4}{(y^2+z^2)(y+z)}+\dfrac{z^4-x^4}{(z^2+x^2)(z+x)}$
$=\dfrac{(x-y)(x+y)(x^2+y^2)}{(x^2+y^2)(x+y)}+\dfrac{(y-z)(y+z)(y^2+z^2)}{(y^2+z^2)(y+z)}+\dfrac{(z-x)(z+x)(z^2+x^2)}{(z^2+x^2)(z+x)}$
$=(x-y)+(y-z)+(z-x)=0$
Suy ra:
$2F=\dfrac{x^4+y^4}{(x^2+y^2)(x+y)}+\dfrac{y^4+z^4}{(y^2+z^2)(y+z)}+\dfrac{z^4+x^4}{(z^2+x^2)(z+x)}$
$\ge\dfrac{(x^2+y^2)^2}{2(x^2+y^2)(x+y)}+\dfrac{(y^2+z^2)^2}{2(y^2+z^2)(y+z)}+\dfrac{(z^2+x^2)^2}{2(z^2+x^2)(z+x)}$
$=\dfrac{x^2+y^2}{2(x+y)}+\dfrac{y^2+z^2}{2(y+z)}+\dfrac{z^2+x^2}{2(z+x)}$
$\ge\dfrac{(x+y)^2}{4(x+y)}+\dfrac{(y+z)^2}{4(y+z)}+\dfrac{(z+x)^2}{4(z+x)}$
$=\dfrac{x+y}{4}+\dfrac{y+z}{4}+\dfrac{z+x}{4}=\dfrac{1}{2}$
$\Rightarrow F\ge\dfrac{1}{4}$
$\min F=\dfrac{1}{4} \Leftrightarrow x=y=z=\dfrac{1}{3}$